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15 tháng 11 2016

Phân tích mẫu thức thành nhân tử :

\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)+b^2c-ab^2+ac^2-bc^2\)

\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)

\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)\)

\(=\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]=\left(b-c\right)\left(a-c\right)\left(a-b\right).\)

Do đó : \(A=\frac{\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3}{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Nhận xét : Nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz.\)

Đặt \(b-c=x,c-a=y,a-b=z\) thì \(x+y+z=0\)

Theo nhận xét trên : \(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3.\)

15 tháng 11 2016

Tử:

(b - c)3 + (c - a)3 + (a - b)3

= (b - c + c - a + a - b)3 - 3(b - c + c - a)(b - c + a - b)(c - a + a - b)

= 0 - 3(b - a)(a - c)(c - b)

= 3(a - b)(a - c)(c - b)

Mẫu:

a2(b - c) + b2(c - a) + c2(a - b)

= a2(b - c) + b2c - ab2 + ac2 - bc2

= a2(b - c) - a(b2 - c2) + bc(b - c)

= a2(b - c) - a(b - c)(b + c) + bc(b - c)

= (b - c)(a2 - ab - ac + bc)

= (b - c)[a(a - b) - c(a - b)]

= (b - c)(a - b)(a - c)

\(A=\frac{3\left(a-b\right)\left(a-c\right)\left(c-b\right)}{\left(b-c\right)\left(a-b\right)\left(a-c\right)}\)

\(=\frac{3\left(c-b\right)}{b-c}\)

3 tháng 11 2018

em ms hok lớp 1

3 tháng 11 2018

Phân tích mẫu \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-c^2b\)

\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b^2-c^2\right)\)

\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b+c\right)\left(b-c\right)\)

\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)=\left(b-c\right)\left[a\left(a-c\right)-b\left(a-c\right)\right]\)

\(=\left(b-c\right)\left(a-b\right)\left(a-c\right)=-\left(b-c\right)\left(a-b\right)\left(c-a\right)\)

Đặt b - c = x, c - a = y, a - b = z

=> x + y + z = b - c + c - a + a - b = 0

Từ x+y+z=0 => x3+y3+z3=3xyz (tự c/m)

=>\(A=\frac{x^3+y^3+z^3}{-xyz}=\frac{3xyz}{-xyz}=-3\)

14 tháng 10 2017

Sửa đề cho nó đẹp

\(\frac{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}\)

\(=\frac{3\left(a-b\right)\left(a-c\right)\left(c-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=-3\)

3 tháng 11 2018

em ms hok lớp 1

14 tháng 10 2018

+) Xét tử thức: \(a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^2\left(a^2-b^2\right)\)

\(=a^3\left(b^2-c^2\right)+\left(b^3c^2-b^2c^3\right)-\left(a^2b^3-a^2c^3\right)\)

\(=a^3\left(b-c\right)\left(b+c\right)+b^2c^2\left(b-c\right)-a^2\left(b-c\right)\left(b^2+bc+c^2\right)\)

\(=\left(b-c\right)\left(a^3b+a^3c+b^2c^2-a^2b^2-a^2bc-a^2c^2\right)\)

\(=\left(b-c\right)\left[\left(a^3b-a^2bc\right)+\left(a^3c-a^2c^2\right)+\left(b^2c^2-a^2b^2\right)\right]\)

\(=\left(b-c\right)\left[a^2b\left(a-c\right)+a^2c\left(a-c\right)-b^2\left(a-c\right)\left(a+c\right)\right]\)

\(=\left(b-c\right)\left(a-c\right)\left(a^2b+a^2c-ab^2-b^2c\right)\)

\(=\left(b-c\right)\left(a-c\right)\left[ab\left(a-b\right)+c\left(a-b\right)\left(a+b\right)\right]\)

\(=\left(b-c\right)\left(a-c\right)\left(a-b\right)\left(ab+bc+ca\right)\)

+) Xét mẫu thức: \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)+b^2c-bc^2-ab^2+ac^2\)

\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)

\(=\left(b-c\right)\left(a^2+bc-ab-ac\right)=\left(b-c\right)\left[\left(a^2-ac\right)-\left(ab-bc\right)\right]\)

\(=\left(b-c\right)\left[a\left(a-c\right)-b\left(a-c\right)\right]=\left(b-c\right)\left(a-c\right)\left(a-b\right)\)

Từ đó; ta có: 

\(\frac{a^3\left(b^2-c^2\right)+b^3\left(c^2-a^2\right)+c^3\left(a^2-b^2\right)}{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}=\frac{\left(b-c\right)\left(a-c\right)\left(a-b\right)\left(ab+bc+ca\right)}{\left(b-c\right)\left(a-c\right)\left(a-b\right)}\)

\(=ab+bc+ca\). KL:...